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Difficulty: Very hardDescriptive Statistics Interpretation

The table below provides operational metrics for 10 international clinical trial sites participating in a multi-center biopharmaceutical study:

Site IDTherapeutic AreaPatients EnrolledDrop-out Rate (%)Average Treatment Duration (Days)
Site AOncology4212.5180
Site BImmunology568.0120
Site COncology3015.0225
Site DNeurology646.2590
Site EOncology7810.0150
Site FImmunology4812.5135
Site GOncology5014.0165
Site HNeurology329.375105
Site IOncology605.0195
Site JImmunology4010.0150

Statement: For the subset of clinical trial sites with a drop-out rate of at least 10.0%, the median Average Treatment Duration exceeds the median Patients Enrolled by more than 110 days.

Answer: Answer

Answer

True. The median Average Treatment Duration for the filtered subset of 6 sites is 157.5 days, and the median Patients Enrolled is 45 patients. The difference of 112.5 days exceeds 110 days.
The statement is correct (True) because filtering for sites with a drop-out rate of at least 10.0%10.0\% yields exactly 6 sites. The median of Patients Enrolled across these 6 sites is 42+482=45\frac{42+48}{2} = 45, and the median of Average Treatment Duration is 150+1652=157.5\frac{150+165}{2} = 157.5. The difference 157.5−45=112.5157.5 - 45 = 112.5 is strictly greater than 110.

Step-by-Step Solution

1
Filter the dataset by the given condition
Subset consists of 6 sites with Drop-out Rate ≥10.0%\ge 10.0\%: Site A, Site C, Site E, Site F, Site G, and Site J.
Only sites meeting the threshold of ≥10.0%\ge 10.0\% drop-out rate must be analyzed.
2
Determine the median of Patients Enrolled for the filtered subset
Ordered values: 30,40,42,48,50,7830, 40, 42, 48, 50, 78. Median = 42+482=45\frac{42 + 48}{2} = 45.
Because the subset size N=6N = 6 is even, the median is the arithmetic mean of the two middle elements (3rd and 4th).
3
Determine the median of Average Treatment Duration for the filtered subset
Ordered values: 135,150,150,165,180,225135, 150, 150, 165, 180, 225. Median = 150+1652=157.5\frac{150 + 165}{2} = 157.5.
Because N=6N = 6 is even, the median is the arithmetic mean of the 3rd and 4th ordered values.
4
Calculate the difference between the two medians and evaluate the statement
Difference = 157.5−45=112.5157.5 - 45 = 112.5 days, which is greater than 110 days.
Since 112.5>110112.5 > 110, the statement is True.

Key Concept

Descriptive Statistics Interpretation on Filtered Even-Count Subsets
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Topics

Descriptive Statistics Interpretation

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